 Calculus III: Two Dimensional Vectors (Level 12 of 13) | Static Equilibrium

Two Dimensional Vectors, level 12.
In this video we will go over how to solve static equilibrium problems. In general when
an object is stationary or at rest we say it is in a state of static equilibrium. When
an object is in this state, the sum of the forces acting on the object must be equal
to 0. For example say we had an object sitting on
a table. In its current position the trophy is not moving but, if the table is suddenly
and rapidly removed, the trophy will fall to the ground and break. What force caused
the trophy to fall down immediately after removing the table? The force responsible
is the force of gravity. This force is perhaps the most intuitive force in our lives since
it pulls us towards the center of the earth on a daily basis. The magnitude of this force
is our weight which is the product of our mass and the gravitational acceleration; the
direction of this force is always towards the center of the earth. Let’s go ahead
and place the trophy back on the table. Since the trophy has a mass it will experience a
gravitational force equal to its mass (m) times the gravitational acceleration (g) pushing
it down towards the center of the earth. Once again this is nothing more than the objects
weight. Now the reason why the trophy does not break the table and move towards the center
of the earth is because the table is pushing the trophy in the opposite direction, and
the magnitude of this force just happens to be the same as the force of gravity exerted
on the trophy, hence the resultant force is equal to zero and the trophy remains motionless. When an object is in a state of static equilibrium
the sum of the forces acting on the object must be equal to 0. In other words the resultant
force equals 0. These types of problem are solved more extensively and in greater detail
in a typical physics course. For this multivariable calculus course we will go over 3 basic examples.
Naveen weighs 42.0 pounds. He is sitting on a swing when his sister Vianey pulls him and
the swing back horizontally through an angle of 30 degrees and then stops. Find the tension
in the ropes of the swing and the magnitude of the force exerted by Vianey.
Alright the figure of this situation is shown below. Notice that Naveen is motionless at
the moment that Vianey stops swinging back. At this point, Vianey is pulling with enough
force to keep Naveen in a state of static equilibrium. In addition to Vianey’s force,
Naveen is under the influence of 2 other forces the gravitational force and the tension force
on the ropes. We are asked to find the magnitude of this tension force and the magnitude of
the force exerted by Vianey. Similar to the way we solved resultant force problems we
need to find the component form of all the forces exerted on the object then we go ahead
and add them component wise and finally we take this sum and set it equal to the zero
vector. Alright let’s go head and first find the component form of all the forces
acting on Naveen. Using the geometry of the diagram we can draw
a right triangle with the ropes representing the hypotenuse and the x and y components
of the tension force representing the legs. For this problem you are free to use the 30
degree angle or the 60 degree angle that is formed by the sides of the right triangle
as long as you apply the trigonometric ratios correctly and assign the correct direction
to the x and y component, everything will work out. I am going to use the 60 degree
angle and use it to find the components of the tension force. In this case the x-component
of the tension force will be equal to the magnitude of the tension force which is currently
not known, times cosine of 60 degrees and the y-component will be equal to the magnitude
of the tension force times sine of 60 degrees. Alright now let’s find the components of
the gravitational force acting on Naveen. Since the gravitational force is always pointing
towards the center of the earth it will only have one component in this case the vertical
or y-component and its magnitude will be equal to Naveen’s weight. Also notice that it
will be negative since it is pointing in the direction of the negative y-axis. So the
components of the gravitational force are going to be equal to 0 for the x-component
and negative 42 pounds for the y-component. Now let’s find an expression for the components
of Vianey’s force, similar to the gravitational force this force will only have one component
which is facing towards the positive x-axis, so the component form of vianey’s force
will be equal to the magnitude of vianey’s pulling force (which is currently not known),
for the x-component and 0 for the y-component. Now that we have all 3 forces we go ahead
and add the forces component wise. Also since this is a static equilibrium problem we also
need to set it equal to the 0 vector as follows. This expression allows us to generate 2 separate
equations one for the x-component and one for the y-component. In essence we are dealing
with a system of equations in this case we have 2 unknowns which require 2 equations
to solve and we just happen to have those equations. So now let’s go ahead and solve
for the tension force with the second equation, approximating the expression we obtain about
48.5 pounds for the tension force. Lastly we use this value and the first equation to
solve and approximate Vianey’s force which is approximately equal to 24.3 pounds. Alright,
let’s try the next example. The tension at each end of the chain has a
magnitude of 25 newtons. What is the weight of the chain? Assume that the weight of the
chain is concentrated at the midpoint of the chain.
Alright here we have a hanging chain under tension forming an angle of 37 degrees with
the horizontal portion of the edges. We are also advised to assume that the weight of
the chain is located at the midpoint of the chain. This means that we are going to have
3 distinct forces acting on the chain focused at the midpoint. Similar to the previous example let’s first
find an expression for the components of all three forces. For the first tension force
the x-component will be equal to the magnitude of the tension force in this case 25 newtons
times cosine of 37 degrees and this component will be negative since it is pointing in the
direction of the negative x-axis. The y-component on the other hand will be equal to 25 newtons
times sine of 37 degrees and it will be pointing towards the positive y-axis. For the second
tension force the magnitudes of both components will be the same, but the x-component will
be pointing towards the positive x-axis. Lastly the gravitational force will be pointing
directly down towards the center of the earth. So its components will be equal to 0 for the
x-component and negative times the magnitude of the weight which is currently unknown.
Next we go ahead and add these 3 vectors and set them equal to the zero vector. Now we
are able to generate our 2 equations by adding the x-components together and the y-components
together, doing that we obtain the following equations. Notice that the first equation
represents the sum of all the x-components acting on the chain and it is composed of
only two forces in this case the x-component of each of the tension force, since this problem
is highly symmetrical we expect these two forces to cancel out and this equation confirms
that observation. The second equation tells us that the entire weight of the chain is
being held by the vertical components of the tension forces, so we go ahead and solve for
the weight, doing that and approximating the expression we obtain 30.1 newtons for the
weight of the chain. Alright let’s go over the final example.
Use the figure below to determine the tension in each cable supporting the given load
Similar to the previous examples let’s start by finding the components of each force, the
first tension force will have the following components, and the second tension force components
will be equal to the following. Finally the gravitational force will have the following
values for its components. Notice that both tension forces are unknowns,
since this problem lacks symmetry we need to get a little bit more involved in order
to find the values of these forces. Let’s go ahead and add these forces and set the
expressions equal to the zero vector. From here let’s extract our two equations
one from each component as follows. Now we are face to face with a system of equations
with 2 unknowns. At this point it is just a matter of solving for one of the unknown
tensions from one equation and substituting it back into the second equation and solving
for the unknown variables. So solving for the first tension force we obtain the following.
Then we go ahead and substitute this expression into the first equation as follows, we proceed
by solving for the magnitude of the second force which is approximately equal to 1958.1
pounds, from here we use this value and solve for the second tension force which is approximately
equal to 2638.2 pounds. Alright in our next video we will go over distance and bearing
problems.

## 4 thoughts on “Calculus III: Two Dimensional Vectors (Level 12 of 13) | Static Equilibrium”

1. KIDGAMING KIDGAMING says:

?

2. KIDGAMING KIDGAMING says:

I do no 12 shit your not my dad you little stupid turtle fu…;-)

3. KIDGAMING KIDGAMING says:

lol👻😂😕 what

4. KIDGAMING KIDGAMING says:

fuck the math coming straight from the underground bitch